1. ## Determinant proof

How do you prove "if det(-A)=0, then det(adj A)=0" ? I am stuck on the right way to prove this. Thank you.

2. Originally Posted by antman
How do you prove "if det(-A)=0, then det(adj A)=0" ? I am stuck on the right way to prove this. Thank you.
Hint 1: $\displaystyle \det(-A) \implies \det (A) = 0$
Hint 2: $\displaystyle A ~\text{adj}(A) = \det (A) I$

3. If det(A)=0, then det(-A) is also 0? I do not know what to do with .

4. Originally Posted by antman
If det(A)=0, then det(-A) is also 0? I do not know what to do with .
Yes, because $\displaystyle \det (-A) = (-1)^n \det (A) = 0$. Since $\displaystyle A~ \text{adj}(A) = I\det (A) = \bold{0}$. If the adjoint was invertible then it would force $\displaystyle A = \bold{0} \implies \text{adj}(A) =\bold{0}$ which is a contradiction. Thus, adjoint is not invertible and so its determinant must be zero.

5. So that proves if det(-A)=0, then det(adj A)=0? and it is indeed true? I am confused about how exactly the adjoint relates to the original matrix, let alone the negative matrix. The adjoint is the transpose of the coefficient matrix, but I am not entirely clear on how to get the coefficient matrix. Would your statement (. Since . If the adjoint was invertible then it would force which is a contradiction. Thus, adjoint is not invertible and so its determinant must be zero.) be a sufficient proof? I am creating an example to go with it. I'm just having a hard time knowing when enough information is given to be considered a proof. Thank you for all of your help!

6. Originally Posted by antman
be a sufficient proof?
Yes. I showed that if $\displaystyle \text{adj}(A)$ is invertible then it would form $\displaystyle \text{adj}(A)$ to be the zero matrix, but that is a contradiction. Therefore, $\displaystyle \text{adj}(A)$ is not invertible, but if it is not invertible then the determinant must be zero.